The function,Originally Posted by c_323_h
is countinous on
For any there exists an such as,
Let and the proof is finished.
Use the Intermediate Value Theorem to show that has a solution.
I understand what the IVT is and how to find a approximate solution using trial and error, but what exactly does it mean by "show?" Are they asking for a proof? I understand the concept of IVT but I can't translate my potential way of solving this into math symbols and such
This is what I tried. Let . Then, by the IVT, there exists a number such thatOriginally Posted by ThePerfectHacker
. Since the function is continuous over the interval there exists a number such that and
I don't know if that proves anything though. I really don't have a direction when I try to prove things. I just copied the definition of IVT out of the book.
Let me try this again.
If is a real function countinous on closed interval , and let be any number between and then there must exist an on such as, .
Root location theorem: If is countinous on and (in other words opposite signs) then there is a solution to the equation on
Proof: By the condition we have that and have opposite signs. Thus, zero is a number between them. Thus, by the intermediate value theorem, for some on
Returning to your problem.
2)The function is countinous cuz all polynomials are.
5)Condition for Root location theorem are satisfied.
6)Thus there exists a solution to .